Question

A balloon initially at rest, starts rising from the ground with an acceleration of $2$ $m/s^2$. After $4$s, a stone is dropped from a balloon. In one second of time after the drop, the stone will cover a distance of $(g=10m/s^2)$.

• A. $3.0$m
• B. $3.4$m
• C. $3.8$m
• D. $2.6$m

Answer 1

Answer: (A)

$3.0$m

Given acceleration of the balloon $a_b$=2 m/$s^2$.

Now velocity of the balloon after t=4 s; $v=u+at$ as initila velocity will be zero.

$v=8m/s$

Now distance covered in 4 seconds $v^2=u^2+2ass=\frac{v^2-u^2}{2a}=16m$ as u=0 m/s

Now as the stone is dropped when the balloon has traveled for 4 s, thus $u_{stone}=v_{balloon}$

So distance traveled by the stone is 1 second $s=ut+\frac{1}{2}g{t}^2=8\left(1\right)-5=3m$

Here g is considered negative as the stone is falling under the action of gravity which is in the direction opposite to that of the velocity of the stone